Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)

On 12/27/24 5:14 AM, WM wrote:

On 26.12.2024 19:41, Jim Burns wrote:

On 12/22/2024 6:32 AM, WM wrote:

 

I (JB) think that it may be that
'almost.all' '(∀)' refers concisely to
the differences in definition at
the center of our discussion.
>
For each finite.cardinal,
almost.all finite.cardinals are larger.
∀j ∈ ℕⁿᵒᵗᐧᵂᴹ: (∀)k ∈ ℕⁿᵒᵗᐧᵂᴹ: j < k

 That is true in potential infinity. It is wrong in actual infinity because there ℕ \ {1, 2, 3, ...} = { } shows that all finite cardinals can be manipulated such that none is larger.

Nope, doesn't show that. It shows that when you have all the finite cardinals, you don't have a "last" one.

>
I think that you (WM) would deny that.
You would say, instead,
ᵂᴹ⎛ for each definable finite.cardinal
ᵂᴹ⎜ almost.all finite cardinals are larger.

 Right.

For sequence ⟨Sₙ⟩ₙ᳹₌₀ of sets
Sₗᵢₘ is a limit.set of ⟨Sₙ⟩ₙ᳹₌₀
if
each x ∈ Sₗᵢₘ is ∈ almost.all Sₖ ∈ ⟨Sₙ⟩ₙ᳹₌₀  and
each y ∉ Sₗᵢₘ is ∉ almost.all Sₖ ∈ ⟨Sₙ⟩ₙ᳹₌₀
⟨Sₙ⟩ₙ᳹₌₀ ⟶ Sₗᵢₘ  ⇐
⎛ x ∈ Sₗᵢₘ  ⇐  (∀)Sₖ ∈ ⟨Sₙ⟩ₙ᳹₌₀:  x ∈ Sₖ
⎝ y ∉ Sₗᵢₘ  ⇐  (∀)Sₖ ∈ ⟨Sₙ⟩ₙ᳹₌₀:  y ∉ Sₖ
>

A limit is a set S​͚  such that nothing fits between it and all sets of the sequence.

---
The notation a​͚  or S​͚  for aₗᵢₘ or Sₗᵢₘ
is tempting, but
it gives the unfortunate impression that
a​͚  and S​͚  are the infinitieth entries of
their respective infinite.sequences.
They aren't infinitieth entries.
They are defined differently.

 The last natural number is finite, and therefore objectively belongs to a finite set. But like all dark numbers it has no FISON and therefore the dark realm appears like an infinite set.

There is no "last" natural numbers, as BY DEFINITION, every natural number has a successor that is bigger than it.
So, your "last" finite number is just a LIE and a figment of your stupidity, so everything you have "proved" based on it is also just a lie and a figment of your stupidity.

>

E(n+2) is
the set of all finite.cardinals > n+2
E(n+1)  =  E(n+2)∪{n+2}
E(n+2)∪{n+2} isn't larger.than E(n+2)

>
Wrong.

>
Almost all finite.cardinals are larger than
finite.cardinal n+1
{n+2} isn't large enough to change that.
Almost all finite.cardinals are larger than
finite cardinal n+2.

 That is true for visible n.

It is true for ALL n, as all n are visible.
Only your stupidity has created the idea of "dark" (or non-visible) numbers, those numbers don't exist in the set you think they do.

>

#E(n+2) isn't any of the finite.cardinals in ℕ

>
It is an infinite number but
even infinite numbers differ like |ℕ| =/= |ℕ| + 1.

>
Infiniteᵂᴹ numbers which differ like |ℕ| ≠ |ℕ| + 1.
are finiteⁿᵒᵗᐧᵂᴹ numbers.

 No. They are invariable numbers like ω and ω+1. The alephs differ only because they count potentially infinite sets which always can be bijected as far as is desired.

No, you just don't understand what you are talking about.

>
The concept of 'limit' is a cornerstone of
calculus and analysis and topology.

 A limit is a number or set such that nothing fits between it and all numbers or sets of the sequence.

Nope. I guess you don't know what a limit actually is.

>
That cornerstone rests upon 'almost.all'.
Each finite.cardinal has infinitely.more
finite.cardinals after it than before it.

 For each finite cardinal that can be defined this is true. Dark cardinals never have been considered.

Dark cardinals don't exist in the finite realm.

 

If one re.defines things away from that,
it is only an odd coincidence if, after that,
any part of calculus or analysis or topology
continues to make sense.
>

Without dark cardinals set theory does not make sense.

Sure it does. It might be needed to make it work in your broken logic system, but that is because your logic is broken.

∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo
and
|ℕ \ {1, 2, 3, ...}| = 0
would contradict each other because more than all n are not in {1, 2, 3, ...}.
 Regards, WM